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otez555 [7]
3 years ago
13

A volume of 80.0 mL of aqueous potassium hydroxide (KOH) is titrated against a standard solution of sulfuric acid (H2SO4). What

was the molarity of the KOH solution if 12.7 mL of 1.50 M H2SO4 was needed
Chemistry
1 answer:
yaroslaw [1]3 years ago
3 0

Answer:

0.478 M

Explanation:

Let's consider the neutralization reaction between KOH and H₂SO₄.

2 KOH + H₂SO₄ → K₂SO₄ + 2 H₂O

12.7 mL of 1.50 M H₂SO₄ react. The reacting moles of H₂SO₄ are:

0.0127 L × 1.50 mol/L = 0.0191 mol

The molar ratio of KOH to H₂SO₄ is 2:1. The reacting moles of KOH are 2 × 0.0191 mol = 0.0382 mol

0.0382 moles of KOH are in 80.0 mL. The molarity of KOH is:

M = 0.0382 mol/0.0800 L = 0.478 M

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7 0
3 years ago
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In an isothermal gas chromatography experiment using an ECD detector, 1.69 nmols of nchlorohexane, C6H13Cl, was added as an inte
prohojiy [21]

Answer:

The amount of Chlorodecane in the unknown is 0.105nmols

Explanation:

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The area of the first peak corresponding to Chlorohexane is 32434 units.

The area of the second peak corresponding to chlorodecane is 2022 units.

Since the response factor of the compound is not given in question and considering the response factor is same for both the compounds, the answer will be as follow:

1.69 nmols of Chlorohexane gives 32434 units

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Answer:

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Therefore, name of the compound will be  Nitrogen triiodide..

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